Question

На доске написаны все натуральные числа от 1 до 12!. К доске подошли девять школьников. Первый умножил все числа на 2, второй умножил полученное число на 3, третий-на ,…, девятый - на 10. Сколько различных остатков при делении на 12! дают полученные числа? Напишите с решением, пожалуйста!

Answer 1

Думаю, что будет $12\cdot 11=132$ разных делителя, потому что остатки просто будут идти с шагом $10!$

Числа $10!, 2\cdot 10!, \ldots, 132\cdot 10!$ дают $132$ разных остатка причем $132*10! = 12!$ дает остаток ноль, а дальше идут числа

$133\cdot 10!=12!+10!=\bmod 12!=10!$ - то есть просто то же самое по второму кругу

Answer 2

Problem Analysis

We are given a scenario where nine students approach a board on which all the natural numbers from 1 to 12! (12 factorial) are written. The first student multiplies all the numbers by 2, the second student multiplies the resulting number by 3, and so on until the ninth student multiplies the number by 10. We need to determine how many different remainders these resulting numbers will have when divided by 12!.

Solution

To solve this problem, we can start by understanding the properties of remainders when dividing by a number.

When a number is divided by another number, the remainder can range from 0 to one less than the divisor. In this case, the divisor is 12!.

To find the number of different remainders, we need to consider the factors of 12!.

Factors of 12!

To find the factors of 12!, we can break it down into its prime factors and their respective powers.

The prime factorization of 12! is: 12! = 2^10 * 3^5 * 5^2 * 7^1 * 11^1

Therefore, the factors of 12! are all the possible combinations of the powers of its prime factors.

Calculating the Number of Different Remainders

To calculate the number of different remainders, we need to consider the factors of 12! and the multipliers used by each student.

The first student multiplies all the numbers by 2, the second student multiplies the resulting number by 3, and so on until the ninth student multiplies the number by 10.

Since the multipliers used by the students are all prime numbers, the resulting numbers will only have the prime factors of 12! and their respective powers.

To find the number of different remainders, we need to consider the powers of the prime factors of 12! and the multipliers used by the students.

Let's calculate the number of different remainders step by step:

1. The first student multiplies all the numbers by 2, so the resulting numbers will have the prime factors of 12! with their powers increased by 1. - The prime factorization of 2 is 2^1. - Therefore, the resulting numbers will have the prime factor 2 with its power increased by 1. 2. The second student multiplies the resulting numbers by 3, so the resulting numbers will have the prime factors of 12! with their powers increased by 1 again. - The prime factorization of 3 is 3^1. - Therefore, the resulting numbers will have the prime factor 3 with its power increased by 1. 3. The third student multiplies the resulting numbers by 5, so the resulting numbers will have the prime factors of 12! with their powers increased by 1 again. - The prime factorization of 5 is 5^1. - Therefore, the resulting numbers will have the prime factor 5 with its power increased by 1. 4. This process continues until the ninth student multiplies the resulting numbers by 10, which is 2 * 5. - The prime factorization of 10 is 2^1 * 5^1. - Therefore, the resulting numbers will have the prime factors 2 and 5 with their powers increased by 1. 5. Finally, we need to calculate the number of different combinations of the powers of the prime factors of 12! after the multiplication process. - The powers of the prime factors of 12! are 2^10, 3^5, 5^2, 7^1, and 11^1. - After the multiplication process, the powers of the prime factors will be increased by 1. - Therefore, the number of different combinations of the powers of the prime factors is the product of (10+1), (5+1), (2+1), (1+1), and (1+1). 6. To find the total number of different remainders, we multiply the number of different combinations of the powers of the prime factors by the number of multipliers used by the students (9 in this case). Let's calculate the number of different remainders using the above steps:

Number of different remainders = (10+1) * (5+1) * (2+1) * (1+1) * (1+1) * 9

The number of different remainders when dividing the resulting numbers by 12! is 6,480.

Conclusion

In this problem, we were given a scenario where nine students multiplied all the numbers from 1 to 12! by different multipliers. We calculated the number of different remainders when dividing the resulting numbers by 12! by considering the factors of 12! and the multipliers used by the students. The number of different remainders is 6,480.